Question

Find the angle between the line `(x-2)/3=(y+3)/3=(z-1)/1` and the plane `2x-3y+4z=0` ?

Answer 1

look carefully, `theta` is the anlge b/w plane and line.

Explanation:

given the direction cosine of line as `vec b = 3hati+3hatj+hatk` and that of plane as `vec n = 2hati-3hatj+4hatk`

clearly , using dot product of vectors,
`cos(90-theta)=hatb.hatn`
Nowit is very easy to find the numerical value of `theta`

Answer 2

`theta ~~ 0.0565" rad "(3.238^@)`

Explanation:

We are given the symmetric form of the equation of the line:

`(x-2)/3=(y+3)/3=(z-1)/1`

Convert the symmetric form of the line to the parametric form by setting each expression equal to t:

`(x-2)/3= t`

`(y+3)/3=t`

`(z-1)/1= t`

Multiply both sides by the denominators:

`x-2= 3t`

`y+3=3t`

`z-1= t`

Move the constants to the right:

`x= 3t+2`

`y=3t-3`

`z= t+1`

Convert to the vector form:

`(x,y,z) = (2,-3,1)+ t(3hati+3hatj+hatk)`

Please understand the `vecu = 3hati+3hatj+hatk` is the direction of the line.

Find any two points in the plane, `2x-3y+4z=0`:

Pick `(0,0,?)`:

`2(0)-3(0)+4z=0`

`z = 0`

The first point is `(0,0,0)`

Pick `(4,4,?)`

`2(4)-3(4)+4z=0`

`z = 1`

The second point is `(4,4,1)`

Make a vector from the first point to the second:

`vecv = (4-0)hati+(4-0)hatj+(1-0)hatk`

`vecv = 4hati+4hatj+1hatk`

Please understand that `vecv` is in the plane and the angle between `vecu` and `vecv` is, also, the angle between the line and the plane.

Compute the dot-product of `vecu` and `vecv`:

`vecu*vecv = 3(4)+3(4)+1(1)=25`

Compute the magnitudes of `vecu` and `vecv`:

`|vecu|= sqrt(3^2+3^2+1^2) = sqrt19`

`|vecv|= sqrt(4^2+4^2+1^2) = sqrt33`

Using the formula `vecu*vecv=|vecu||vecv|cos(theta)`, we can find the angle between the two vectors:

`25 = sqrt19sqrt33cos(theta)`

`theta = cos^-1(25/(sqrt19sqrt33))`

`theta ~~ 0.0565" rad "(3.238^@)`